Question

Two tangent segments BC and BD are drawn to a circle with centre O such that ∠CBD = 120°. Prove that OB = 2BC.

Answer 1

Given: Two tangent segments BC and BD are drawn from an external point B to a circle with centre O and points of contact C and D and ∠CBD = 120°.

To Prove: OB = 2BC.

Proof [Step-wise]:

1. BC = BD because tangents drawn from the same external point to a circle have equal lengths.

2. The angle between the two tangents ∠CBD = 120° is supplementary to the central angle subtended by the points of contact C and D; hence ∠COD = 180° – 120° = 60°. (Theorem: angle between two tangents + angle subtended at centre = 180°).

3. OC = OD = r (radii). In isosceles triangle COD the included angle ∠COD = 60°, so by the Law of Cosines or recognizing an equilateral triangle CD = r.

Calculation (Law of Cosines): 

CD² = r² + r² – 2r² 

`cos 60^circ = 2r^2 - 2r^2(1/2) = r^2` 

⇒ CD = r

4. Consider triangle BCD. It is isosceles with BC = BD = t (say) and vertex angle at B equal to 120°.

Use the Law of Cosines to find CD in terms of t: 

CD² = t² + t² – 2t²

`cos 120^circ = 2t^2 - 2t^2(-1/2) = 3t^2` 

⇒ `CD = tsqrt(3)`

5. Equate the two expressions for CD from steps 3 and 4:

`r = tsqrt(3)` 

⇒ `t = r/sqrt(3)`. 

Thus `BC = r/sqrt(3)`.

6. In right triangle OCB (OC ⟂ BC because radius to point of contact is perpendicular to tangent).

OC = r, BC = t = `r/sqrt(3)`

So OB² = OC² + BC² 

= `r^2 + (r^2/3)`

= `(4/3)r^2` 

Hence `OB = (2/sqrt(3))r`.

7. Since `BC = r/sqrt(3)`

`2 xx BC = 2 xx (r/sqrt(3))` 

= `(2/sqrt(3))r` 

= OB

Therefore OB = 2BC.